Thread: Checkmate!

How do you put 8 queens on a chessboard so that they won't capture each other?

Being on a programming forum and all, you could write a program that finds the answer for you...
Or look below:

This one actually works. (There are 92 working solutions)

You could create a program to find the solution using backtracking.... Similar to solving a maze problem.. I have to use something like this when i start coding my chess games single player part..

Originally posted by minesweeper
Won't the top and bottom-most queens capture each other?

Gee, how could I miss that one???
And I thought I checked em all....

And if you want how I calculated it, here ya go.

Warning
This code is uncommented, and basically contest-style code.

Code:

#pragma warning( disable: 4530 )

#include <fstream>

using namespace std;

#define MAXSIZE 13


bool col[ MAXSIZE ];
bool updiag[ MAXSIZE * 2 ];
bool downdiag[ MAXSIZE * 2 ];

ifstream in("checker.in");
ofstream out("checker.out");

int n = 0;
int numoutputed = 0;

void outputsoln( int *row ) 
{
  
  for( int i = 0; i < n; i++ ) {
    
    if( i != 0 ) out<<" ";
    
    out<<row[i] + 1;
    
  }

  out<<endl;
  
}

int calcnum( int i, int *row ) 
{
  int sum = 0;
  
  if( i == n ) {
    
    outputsoln( row );
    return 1;
    
  }
  
  for( int j = 0; j < n; j++ ) {
    
    if( !col[ j ] && !updiag[ i + j ] && !downdiag[ i - j + n ] ) {
      
      col[ j ] = true;
      updiag[ i + j ] = true;
      downdiag[ i - j + n ] = true;
      row[i] = j;
    
      sum += calcnum( i + 1, row );
    
      col[ j ] = false;
      updiag[ i + j ] = false;
      downdiag[ i - j + n ] = false;
      
    }
    
  }

  return sum;
  
}


int main( void ) 
{
  int rows[ MAXSIZE ];
  
  in>>n;
  
  out<<calcnum( 0, rows )<<endl;
  
  
  return 0;
}